Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 10

CIE/GCE/AS/Math/P1/19/Nov/13/Q#10

Question

Relative to an origin O, the position vectors of points A, B and X are given by


−8 10 −2
⃗⃗⃗⃗⃗ = (−4)
𝑂𝐴 ⃗⃗⃗⃗⃗ = ( 2 )
𝑂𝐵 and ⃗⃗⃗⃗⃗ = (−2)
𝑂𝑋
2 11 5


i. Find ⃗⃗⃗⃗⃗
𝐴𝑋 and show that AXB is a straight line.

1
⃗⃗⃗⃗⃗
The position vector of point C is given by 𝑂𝐶 = (−8).
3

ii. Show that CX is perpendicular to AX.

iii. Find the area of triangle ABC.


Solution

i.

⃗⃗⃗⃗⃗ .
First, we are required to find 𝐴𝑋
A vector in the direction of ⃗⃗⃗⃗⃗
𝐴𝐵 is;

⃗⃗⃗⃗⃗
𝐴𝐵 = 𝐛 − 𝐚 = ⃗⃗⃗⃗⃗
𝑂𝐵 − ⃗⃗⃗⃗⃗
𝑂𝐴

For the given case;

⃗⃗⃗⃗⃗
𝐴𝑋 = 𝐱 − 𝐚 = ⃗⃗⃗⃗⃗
𝑂𝑋 − ⃗⃗⃗⃗⃗
𝑂𝐴

We are given that;

−8
⃗⃗⃗⃗⃗ = (−4)
𝑂𝐴
2

, CIE/GCE/AS/Math/P1/19/Nov/13/Q#10
−2
⃗⃗⃗⃗⃗
𝑂𝑋 = (−2)
5



−2 −8 −2 + 8 6
⃗⃗⃗⃗⃗
𝐴𝑋 = 𝐱 − 𝐚 = (−2) − (−4) = (−2 + 4) = (2)
5 2 5−2 3


6
⃗⃗⃗⃗⃗
𝐴𝑋 = (2) = 6𝐢 + 2𝐣 + 3𝐤
3


Next, we are required to show that AXB is a straight line.

It is evident that if AXB is straight line then angle between AX and BX must be 180𝑜.

Let’s first find BX and then angle between AX and BX.

A vector in the direction of ⃗⃗⃗⃗⃗
𝐴𝐵 is;

⃗⃗⃗⃗⃗ = 𝐛 − 𝐚 = 𝑂𝐵
𝐴𝐵 ⃗⃗⃗⃗⃗ − 𝑂𝐴
⃗⃗⃗⃗⃗

For the given case;

⃗⃗⃗⃗⃗ = 𝐱 − 𝐛 = 𝑂𝑋
𝐵𝑋 ⃗⃗⃗⃗⃗ − 𝑂𝐵
⃗⃗⃗⃗⃗

We are given that;

10
⃗⃗⃗⃗⃗
𝑂𝐵 = ( 2 )
11
−2
⃗⃗⃗⃗⃗
𝑂𝑋 = (−2)
5